Introduction to Aerospace Engineering

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1 Introduction to Aerospace Engineering Lecture slides Challenge the future

2 4-0-0 Introduction to Aerospace Engineering Aerodynamics 3 & 4 Prof. H. Bijl ir. N. Timmer Delft University of Technology Challenge the future

3 3 & 4. Anderson Ernst Mach

4 Picture on front page: First picture ever of the shock wave around a bullet at supersonic speed presented by Mach at the Academy of Sciences in Vienna in 887 The two vertical lines are made by the trip wires that triggered the camera as the projectile passed by He employed an innovative approach called the shadowgraph (predecessor of the Schlieren technique) based on the diffraction of light that changes with air density and temperature 3

5 Subjects lecture 3 Thermodynamics - First law of thermodynamics - Specific heat -Equations for a perfect gas Isentropic flow Energy equation 4

6 boundary surroundings System (Unit mass of gas) e is internal energy of the system δq (heat added) 5

7 e can only change when: Heat is added or taken away from the system (δq is heat per unit mass) Work is done on, or by, the system (δw is work done per unit mass) de= δq+ δw First law of thermodynamics 6

8 Different forms of first law For the work at surface: work is Pressure x Areax distance= p da s p da s δ w = pdas = p sda = pdv A A 7

9 For a reversible process : δw=-pdv dv is an incremental change in the volume due to a displacement of the boundary of the system hence : de= δq-pdv 8

10 Enthalpy Lets introduce another quantity of the flow: the specific enthalpy h h=e + pv With differential: dh=de+d(pv) dh=de+pdv+vdp dh= δq + vdp 9

11 Types of Processes st example δq p,t,v Constant Volume Process rigid boundary = constant volume 0

12 nd example Constant Pressure Process p,t,v δq Piston moves just to keep pressure p constant

13 Specific Heat Definition of Specific Heat : (depends on type of process, thus=multivalued) c δq dt The resulting value of dt when adding δq depends on the type of process! c c v δq = dt δq = dt at constant volume p at constant pressure Now relate c v and c p to e, h etc.

14 Specific Heat δ q = de + pdv = de For constant volume process dv = 0 de = c v dt This can be integrated to : e = c v T For constant pressure process dp = 0 q dh vdp de δ = = dh = c p dt This can be integrated to : h = c p T (h=0 at T=0) Above relations also hold in general as long as the gas is a perfect gas. 3

15 Equations for a perfect gas Hence for a perfect gas (no molecular forces) : de = c v dt dh=c p dt e=c v T For air: c v = 70 J/(kg K) c p = 008 J/(kg K) h=c p T (For any process with T < 600 K) 4

16 Isentropic flow We introduce one more concept to bridge thermodynamics and compressible aerodynamics : Isentropic flow Definitions: Adiabatic process : δq = 0 Reversible process : no frictional or dissipative effects Isentropic Process : both adiabatic and reversible. Note: Though the flow can be isentropic the TEMPERATURE might change from point to point! 5

17 Isentropic flow, ratio of specific heats For isentropic flow (or a perfect gas) Several important relationships between p, T, ρ: Since δq = 0 δq = de + pdv = 0 Since δq = 0 δq = dh - vdp = 0 -pdv = de = c v dt vdp = dh = c p dt -pdv vdp = c c v p dp p c p = - c v dv v dp p dv = γ v Ratio of specific heats c c p v = γ =.4 for air 6

18 Isentropic Flow Now integrate between points and on a streamline... dp = - γ p dv v [ ] ln p ln p = γ ln v ln v Remember : p v ln = - γ ln p v ln(a) - ln(b) = ln a b [ ln p] γ [ ln v] and p v = p v = a ln b = ln(b a ) -γ 7

19 Since the specific volume is v ρ v = = we find : ρ p p -γ ρ p ρ = = ρ p ρ γ for ISENTROPIC FLOW 8

20 Isentropic Flow Now with the equation of state pv=rt or p/ρ=rt we may write : γ γ -γ γ p p RT p T p T = = = p RT p p T p T -γ -γ p T p T = = γ p T p T γ - for ISENTROPIC FLOW AGAIN: Remember Isentropic flow relations are only relevant to compressible flow 9

21 Isentropic Flow Equations p p = ρ ρ γ = T T γ /(γ ) 0

22 Energy equation for frictionless adiabatic flow, derived from the st law of thermodynamics Physical principle: Energy can neither be created nor destroyed

23 Energy equation.. (ctnd) One of the forms of the st law of thermodynamics: δq = dh - vdp For adiabatic flow δ q= 0 dh - vdp = 0

24 Energy equation.. (ctnd) dh - vdp = 0 For frictionless flow the Euler equation gives: dp = - ρ VdV With this the st law becomes: 0=dh + v ρ VdV 3

25 Energy equation.. (ctnd) Since v=/ρ(= specific volume) we find: dh + VdV = 0 Now integrate between two points along a streamline... 4

26 Energy equation for frictionless, adiabatic flow h v h dh + VdV = 0 v h + V = h + V h - h+ ( V - V ) = 0 or h + V = constant Since also h = C p T CpT + V = constant 5

27 Summary of equations For steady, frictionless, incompressible flow: Continuity equation A V = A V Bernoulli s equation p + ρ V = p + ρ V 6

28 For steady, isentropic compressible flow (adiabatic and frictionless) : Continuity equation ρ A V = ρ A V Isentropic relations Energy equation γ p ρ γ = = T γ- p ρ T c p T + V = c p T + V Equation of state P = ρ RT P = ρ RT 7

29 Example 4.9: A supersonic wind tunnel is considered. The air temperature and pressure in the reservoir of the tunnel are: T 0 = 000 K and p 0 = 0 atm. The static temperatures at the throat and exit are: T* = 833 K and T e = 300 K. The mass flow through the tunnel is 0.5 kg/s. For air, c p = 008 J/(kg K). Calculate: a) Velocity at the throat V* b) Velocity at the exit V e c) Area of the throat A* d) Area of the exit A e 8

30 Subjects lecture 4 Speed of sound Measurement of air speed for: - incompressible, frictionless flow -subsonic, high speed adiabatic, frictionless flow 9

31 Summary of equations For steady, frictionless, incompressible flow: + p A V = A V ρv =p V + ρ Continuity equation Bernoulli s equation 30

32 For steady, isentropic (adiabatic and frictionless) compressible flow: c p T ρ A V = ρ A V γ γ p ρ = = T γ- p ρ T + V =c p T + V Continuity equation Isentropic relations Energy equation P = ρrt Equation of state 3

33 Speed of sound p p+dp p p+dp a ρ T ρ +dρ T+dT is equal to ρ T a ρ +dρ T+dT a + da moving sound wave with speed a into a stagnant gas motionless sound wave static observer observer moving with sound wave 3

34 Speed of sound Apply the continuity equation: ρ A V = ρ A V ρa a = (ρ + dρ) A (a+da) -dimensional flow A = A = A = constant Thus: ρa = (ρ + dρ)(a + da) ρa = ρa + adρ + ρda + dρda da a = - ρ d ρ small, ignore 33

35 Speed of sound Now apply momentum equation (Euler equation) dp = - ρvdv Substituting this into We obtain a = dp dρ dp = - ρada da = - dp ρa da a = - ρ d ρ dp a = + ρ ρ a d ρ Going through the sound wave there is no heat addition, friction is negligible a = dp dρ isentropic 34

36 Speed of sound Now we may apply the isentropic relations. We want to rewrite dp dρ With isentropic relations we find a = p γ ρ Use the equation of state : p = RT ρ a = γ RT Speed of sound in a perfect gas depends only on T! 35

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38 a = γ RT Mach M = V a Ackeret In honor of Ernst Mach the name Mach number was introduced in 99 by Jacob Ackeret 37

39 A380: 005 present Max # pass. 853 Mach=

40 A380 flight deck 39

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44 BAC-Sud Aviation CONCORDE Mach=.04 Ticket London New York,000 # pass. 8 43

45 Flight deck of Concorde 44

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49 Tupolev TU-44 48

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51 Equations for a perfect gas We defined: h=e + pv We found : h=c p T and e= c v T With the eq. of state: pv=rt we find: c p T=c v T+ RT hence: c p =c v + R => c p -c v = R We defined: c γ = c v p R c p = γ γ 50

52 Measurement of airspeed /flow from a reservoir : V=0 Energy equation: c pt + V = constant c T + V = c T o + V p p o Assume index 0 = stagnation point V o = 0 Substitute: T o V c p T + V = c p T o = + T c T C p p γ R = (from Cp Cv = R ) γ T o γ - V = + T γ RT 5

53 Measurement of airspeed /flow from a reservoir : V=0 With a = γ RT T γ γ = + = + M T a o - V - Bring flow isentropically to rest Isentropic relation can be used p γ γ - o ρ o T o γ = = ρ T p po γ - = + M p γ γ - ρ o γ - = + M ρ γ - 5

54 Second form of isentropic relations T γ- = + M T o po γ- = + M p γ γ- ρo ρ γ - = + M γ - 53

55 ρ ρ ρ ρ 0 γ - = + M γ Derived from energy equation and isentropic relations M For M < 0.3 the change in density is less than 5 % Thus : for M < 0.3 the flow can be treated as incompressible 54

56 Nozzle flow flow M T/T 0 P/P 0 ρ/ρ 0 55

57 Measurement of airspeed, subsonic low speed (=incompressible), frictionless flow Subsonic nozzle: p,v,a P,V,A P 3,V 3,A 3 What is value of the air speed V? P total :V = 0 56

58 Measurement of airspeed, incompressible, frictionless flow Bernoulli: p+ ρ V = p + ρ V ρ V = p - p + ρ V V = A Continuity V A = V A V A substitute = p - p + ρ V ρ A V A V = ( p - p ) ρ - A A - A ρ ρ = p - p A V When pressure difference p -p is measured, V is known 57

59 Measurement of airspeed, supersonic flow M > Formation of a shock wave : Across the shock wave: M> M > M p < p Flow through shock-wave is : V > V T < T non-isentropic p t > p t T t = T t (for perfect gas) Because the flow is non-isentropic a special shock wave theory must be developed to relate the pitot tube measurement to M. ( This is beyond the scope of our lecture!). 58

60 Measurement of airspeed, supersonic flow M > presence is not communicated to the flow upstream shock wave (large friction and thermal conduction!) 59

61 Opgave In de verbrandingskamer van een raketmotor worden kerosine en zuurstof verbrand. Dit resulteert in een heet gasmengsel onder hoge druk in de verbrandingskamer met de volgende condities en eigenschappen: T 0 = 500 K p 0 = 8 atm R = 378 J/kg K γ=.6 De druk in de uitlaat van de tuit van de raketmotor is atm, en de oppervlakte van de keel is gelijk aan 0.08 m. Neem isentrope stroming aan. Gevraagd: Bereken de snelheid in de uitlaat van de raketmotor. Bereken de massastroom door de tuit van de raketmotor. verbrandings kamer keel uitlaat 0 tuit 60

62 Problem In the combustion chamber of a rocket engine, kerosene is burned, resulting in a hot, high-pressure gas mixture with the following properties: T 0 = 500 K p 0 = 8 atm R = 378 J/kg K γ=.6 This gas flows from the combustion chamber through the rocket nozzle. The pressure at the exit of the rocket nozzle is: atm.andthe area of the throat of the rocket nozzle is: 0.08 m. Assume isentropic flow. Compute the velocity at the exit of the rocket nozzle. Compute the mass flow through the rocket nozzle. combustion chamber throat exit 0 nozzle 6

Introduction to Aerospace Engineering

Introduction to Aerospace Engineering Lecture slides Challenge the future 3-0-0 Introduction to Aerospace Engineering Aerodynamics 5 & 6 Prof. H. Bijl ir. N. Timmer Delft University of Technology 5. Compressibility